A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0 . 05 . When hit by a bullet of mass 50 g moving with speed v , that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed v 10 after being dropped from height H , then neglecting energy losses and taking g = 10 m s - 2 , the value of H is close to

Question

A block of mass m = 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0 . 05 . When hit by a bullet of mass 50 g moving with speed v , that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed v 10 after being dropped from height H , then neglecting energy losses and taking g = 10 m s - 2 , the value of H is close to, 0 . 2 km ., 0 . 5 km ., 0 . 4 km ., None of these.

MARKS App · Accepted Answer

By the conservation of linear momentum, m 1 v = m + m 1 v 1 ⇒ v 1 = m 1 v m + m 1 . . . ( 1 ) After collision by work energy theorem, we have W friction = ∆ k ⇒ - μ m + m 1 g x = - 1 2 m + m 1 v 1 2 ⇒ μ g x = 1 2 v 1 2 ⇒ 0.05 × 10 × 2 = 1 2 m 1 v m + m 1 2 ⇒ 5 100 × 10 × 4 = 50 1000 2 × v 2 10 + 50 1000 2 ⇒ 2 = 1 400 v 2 × 400 201 2 v 2 = 201 2 2 v = 2 × 201 For a freely falling body, v ' = 2 g H v 10 = 2 g H 2 × 201 10 = 2 × g H 201 2 100 = 10 H H = 201 2 1000 = 40 m = 0.04 km

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